change variable values based on preceding value - r

Question

I have the following dataset:
df <- data.frame(subject = c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3),
time = c(1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,11),
performance = c(1,0,-1,-1,0,1,1,-1,0,0,0,1,1,1,-1,0,1,1,-1,0,0,1,-1,1,1,0,1,1,-1,0,-1,-1,0))
What I would like to do is to change some of the entries in the performance variable. More specifically, if a "-1" entry is preceded by a "1", I want to change the "-1" to "0".
However, this should be done within subjects only, but not across subjects (all of the subjects have a varying number of sessions).
So, this is what I'd like to have in the end:
df2 =data.frame(subject = c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3),
time = c(1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,11),
performance = c(1,0,-1,-1,0,1,1,0,0,0,0,1,1,1,0,0,1,1,0,0,0,1,-1,1,1,0,1,1,-1,0,-1,-1,0))
Does anyone have an idea how to do this?
Thanks in advance!
S.

Answer 1

Using dplyr,
df %>%
group_by(subject) %>%
mutate(performance = replace(performance, which(performance + lag(performance)==0 & performance == -1), 0))

Answer 2

Here's a data.table approach, where I first create a flag column which is then used to subset the data and update the performance column by reference.
library(data.table)
dt <- as.data.table(df) # or setDT(df)
dt[, flag := performance == -1 & shift(performance, 1L) == 1, by = subject]
dt[(flag), performance := 0][, flag := NULL]
I chose to do it with an intermediate flag-column because I expect that to perform very well for large data sets. If performance is not your concern, you could of course use ifelse or replace instead.

Answer 3

This is ugly, but should work:
dftest <- df
for (i in 2:nrow(dftest)) {
if(
dftest$performance[i] == -1 && dftest$performance[i - 1] == 1
){
if(
dftest$subject[i] == dftest$subject[i - 1]
) {
dftest$performance[i] <- 0
}
}
}
all.equal(df2, dftest) # ONE ERROR
This gives an error in line 29 - can you check whether your example df2 is correct here? If I understand the question correctly df2$performance[29] should be 0?

Answer 4

A base R solution using by and sapply:
gr <- do.call(c, by(df, df$subject, function(x) {
c(FALSE, unlist(sapply(1:length(x$performance),
function(y) (x$performance[y] == -1) & (x$performance[y-1] == 1))))
}))
df[gr, 3] <- 0
cbind(df, df2)